Question

help me with this question​

Answer 1

$\large\text{\underline{Note}}$

For your question to be solvable, it should require a positive solution only, so I am rejecting the 'negative $x$ case'.

$\large\text{\underline{Solution}}$

We are given that,

$\implies x^{2}=7+4\sqrt{3}$

We can observe that,

$\implies x^{2}=(2+\sqrt{3})^{2}$

That is,

$\implies x=2+\sqrt{3}\text{ (Accept.)}\text{ or }x=-2-\sqrt{3}\text{ (Reject.)}$

If we complete the square of $x^{6}-52x^{3}+1$,

$\implies x^{6}-52x^{3}+1=(x^{6}-52x^{3}+26^{2})-26^{2}+1$

$\implies x^{6}-52x^{3}+1=(x^{3}-26)^{2}-26^{2}+1\text{ ... (A)}$

However,

$\implies x^{3}-26=(2+\sqrt{3})^{3}-26$

$\implies x^{3}-26=(8+12\sqrt{3}+18+3\sqrt{3})-26$

$\implies x^{3}-26=-15\sqrt{3}$

Hence,

$\implies(x^{3}-26)^{2}=675\text{ ... (B)}$

By substituting B into A,

$\implies x^{6}-52x^{3}+1=675-26^{2}+1$

$\implies x^{6}-52x^{3}+1=675-676+1$

$\implies x^{6}-52x^{3}+1=0$

$\large\text{\underline{Conclusion}}$

Hence, the value is 0 for the positive value of $x$.

$\large\text{\underline{Footnote}}$

If you need the case of negative $x$, the required value will come as $2704+1560\sqrt{3}$, so the negative solution gets rejected. Thank you.

Answer 2

Solution :-

→ x² = 7 + 4√3

→ x² = (4 + 3 + 2 * 2 * √3)

→ x² = (2)² + (√3)² + 2 * 2 * √3

→ x² = (2 + √3)²

→ x = (2 + √3)

so,

→ 1/x = 1/(2 + √3) * {(2 - √3)/(2 - √3)} = (2 - √3) / (2)² - (√3)² = (2 - √3)

then,

→ x + 1/x = (2 + √3) + (2 - √3) = 4

→ (x + 1/x)³ = 4³

→ x³ + 1/x³ + 3 * x * 1/x * (x + 1/x) = 64

→ (x³ + 1/x³) + 12 = 64

→ (x³ + 1/x³) = 64 - 12

→ (x³ + 1/x³) = 52

therefore,

→ x⁶ - 52x³ + 1

→ x³(x³ - 52 + 1/x³)

→ x³[(x³ + 1/x³) - 52]

→ x³[52 - 52]

→ x³ * 0

→ 0 (Ans.)

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if a²+ab+b²=25

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